Math Doubts

Find $x^2+y^2+z^2$ if $x = r\cos{\alpha}\cos{\beta}$, $y = r\cos{\alpha}\sin{\beta}$, and $z = r\sin{\alpha}$

$x^2+y^2+z^2$ is an algebraic expression. It is given that the values of all three literals are expressed in three equations as follows.

$(1) \,\,\,\,\,\,$ $x = r\cos{\alpha}\cos{\beta}$

$(2) \,\,\,\,\,\,$ $y = r\cos{\alpha}\sin{\beta}$

$(3) \,\,\,\,\,\,$ $z = r\sin{\alpha}$

It is asked us to find the value $x^2+y^2+z^2$ in this problem on the basis of the above three equations.

Substitute values of x, y and z in the expression

Replace the algebraic expression by the respective values of the literals $x$, $y$ and $z$ to find the value of algebraic expression.

$x^2+y^2+z^2$ $\,=\,$ ${(r\cos{\alpha}\cos{\beta})}^2$ $+$ ${(r\cos{\alpha}\sin{\beta})}^2$ $+$ ${(r\sin{\alpha})}$

$=\,\,\,$ $r^2\cos^2{\alpha}\cos^2{\beta}$ $+$ $r^2\cos^2{\alpha}\sin^2{\beta}$ $+$ $r^2\sin^2{\alpha}$

Take common factors out from the terms

Observe the first two terms in the expression, $r^2\cos^2{\alpha}$ is a common factor in both terms in the expression. So, they can be taken common from them. It is not only useful to simplify the expression and helpful to us to find the value of the expression easily.

$=\,\,\,$ $r^2\cos^2{\alpha}{(\cos^2{\beta}+\sin^2{\beta})}$ $+$ $r^2\sin^2{\alpha}$

Simplify the trigonometric expression

According to Pythagorean identity of sin and cos functions, the value of $\cos^2{\beta}+\sin^2{\beta}$ is equal to one.

$=\,\,\,$ $r^2\cos^2{\alpha}{(1)}$ $+$ $r^2\sin^2{\alpha}$

$=\,\,\,$ $r^2\cos^2{\alpha}$ $+$ $r^2\sin^2{\alpha}$

Continue simplifying the expression

This time, $r^2$ is a common factor in the two terms of the expression. So, take it common from them to proceed further in simplifying the trigonometric expression.

$=\,\,\,$ $r^2{(\cos^2{\alpha}+\sin^2{\alpha})}$

Now, use Pythagorean identity of sin and cos functions to get the complete solution of this problem.

$=\,\,\,$ $r^2{(1)}$

$=\,\,\,$ $r^2$

Therefore, it is solved that the value of $x^2+y^2+z^2$ is equal to $r^2$.

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